I realize your "base case" can be with a negative value, ex. if $P(-5)$ is true and $P(k+1)$ is true given that $P(k)$ is true, then $P(x)$ is true for all $x|x\in \mathbb{R},x>-5$
But that does not prove that proposition $P$ is true for any $x \leq 5$
If any of the above is incorrect please let me know.
My question is whether induction can go in the opposite direction. For example, proving that $P(k-1)$ is true if $P(k)$ is true, and proving that, say, $P(5)$ is true, would let you say conclusively that proposition $P$ is true for $x \leq 5$. Is this valid, and if so does it have a particular name (seeing as I always see induction defined only on the natural numbers)?
Sure, the process will work similarly to the standard induction setup.
Take a simple example:
$$\forall n \le 2 , 2n+1 \le 5$$
The base case is easy enough to see. Then notice that if we assume the statement is true for $k$ i.e. $2k+1 \le 5$ then we have for $(k-1)$ that $$2(k-1) +1 = 2k+1 -2 \le 5-2 \lt 5$$
So what did we show? We've shown that this statement is true for $n= 2$ and we shown that if the statement is true for some number $k$, then it will be true for $k-1$. Combining these two results let's us complete our proof.